Problem 91 - Calculus for Biology and Medicine

Question

\(X\) and \(Y\) are independent and Poisson with mean 3 . (a) Find \(P(X+Y=2)\). (b) Given that \(X+Y=2\), find the probability that \(X=k\) for \(k=0,1\), and \(2 .\)

Step-by-Step Solution

Answer

(a) \(P(X+Y=2)=18e^{-6}\). (b) \(P(X=k|X+Y=2)=\frac{1}{12},\frac{1}{3},\frac{1}{4}\) for \(k=0,1,2\).

Step 1: Identify Distribution Properties

We know that both \(X\) and \(Y\) are independent Poisson random variables with a mean of 3. For a Poisson random variable, the probability of observing \(k\) events is given by \(P(Z=k) = \frac{e^{- ext{mean}} imes ext{mean}^k}{k!}\). Thus, \(X\sim \text{Poisson}(3)\) and \(Y\sim \text{Poisson}(3)\).

Step 2: Sum of Independent Poisson Variables

For independent Poisson random variables, the sum \(X+Y\) is also Poisson distributed. The mean of \(X+Y\) is the sum of the means, so \(X+Y\sim \text{Poisson}(6)\) because both individual means are 3.

Step 3: Calculate Probability for \(X+Y=2\)

Using the Poisson probability mass function for \(X+Y\), we find \(P(X+Y=2) = \frac{e^{-6} imes 6^2}{2!}\). Simplifying, \(P(X+Y=2) = \frac{e^{-6} imes 36}{2} = 18e^{-6}\).

Step 4: Understand Conditional Probability Setup

Given \(X+Y=2\), we need to find \(P(X=k|X+Y=2)\). By the definition of conditional probability, \(P(X=k|X+Y=2) = \frac{P(X=k \text{ and } Y=2-k)}{P(X+Y=2)}\).

Step 5: Calculate Probabilities for \(P(X=k, Y=2-k)\)

To find the joint probability \(P(X=k, Y=2-k)\), calculate the individual Poisson probabilities: \(P(X=k) = \frac{e^{-3} imes 3^k}{k!}\) and \(P(Y=2-k) = \frac{e^{-3} imes 3^{2-k}}{(2-k)!}\). Multiply these probabilities since X and Y are independent.

Step 6: Calculate Conditional Probabilities for \(k=0,1,2\)

Compute \(P(X=k, Y=2-k)\) for each \(k\): - For \(k=0\), \(P(X=0, Y=2) = \frac{e^{-3} imes 3^0}{0!} \times \frac{e^{-3} imes 3^2}{2!}\) - For \(k=1\), \(P(X=1, Y=1) = \frac{e^{-3} imes 3^1}{1!} \times \frac{e^{-3} imes 3^1}{1!}\) - For \(k=2\), \(P(X=2, Y=0) = \frac{e^{-3} imes 3^2}{2!} \times \frac{e^{-3} imes 3^0}{0!}\)\(P(X=k|X+Y=2)\) is calculated by dividing each of these by \(P(X+Y=2) = 18e^{-6}\).

Step 7: Derive Final Conditional Probabilities

Compute the final probabilities: - \(P(X=0|X+Y=2) = \frac{1}{12}\) - \(P(X=1|X+Y=2) = \frac{1}{3}\) - \(P(X=2|X+Y=2) = \frac{1}{4}\)

Key Concepts

Poisson Distribution

The Poisson distribution is a probability distribution that describes the number of events happening within a fixed interval of time or space. This is used when events occur independently of each other and the average rate at which the events happen is constant over time. It is characterized by a single parameter, which is the mean number of occurrences (c). This mean indicates how frequently the events take place. In the exercise, both variables, \(X\) and \(Y\), each follow a Poisson distribution with a mean of 3.

The probability of obtaining exactly \(k\) events in a Poisson distribution is given by:
\[ P(Z=k) = \frac{e^{-\lambda} \times \lambda^k}{k!} \]
where:

• \(\lambda\) is the average number of events (mean).
• \(k!\) is the factorial of \(k\).
• \(e\) is the base of the natural logarithm, approximately equal to 2.718.

This formula provides the probability of observing \(k\) events given \(\lambda\). Thus, for \(X\sim \text{Poisson}(3)\), the mean is 3, which indicates events frequently occur every cycle on average at this rate.

Independent Variables

When we talk about independent variables in probability, we mean that the occurrence of one event does not affect the probability of the occurrence of another. This property is crucial when dealing with the sum of random variables in distributions like Poisson.

The exercise involves two independent Poisson random variables \(X\) and \(Y\). This independence allows us to combine their distributions. Specifically, the sum of independent Poisson variables results in another Poisson distribution. The mean of this new distribution is simply the sum of the individual means.

• For this exercise:
  \(X \sim \text{Poisson}(3)\) and \(Y \sim \text{Poisson}(3)\).
• Therefore, the sum \(X + Y \sim \text{Poisson}(6)\).

This property makes it straightforward to derive the probability distribution of the variables' sum. Thus, to find probabilities such as \(P(X+Y=2)\), you use the new Poisson distribution with mean 6, calculated as shown in the problem solution.

Conditional Probability

Conditional probability refers to the probability of an event occurring given that another event has already occurred. It is calculated by taking the probability of both events happening together and dividing it by the probability of the given (or "conditioning") event.

In the provided exercise, given that \(X+Y=2\), we are asked to find the probability that \(X=k\) for \(k = 0, 1, 2\). The conditional probability is expressed as:

\[ P(X=k|X+Y=2) = \frac{P(X=k \text{ and } Y=2-k)}{P(X+Y=2)} \]

This formula helps in understanding the likelihood of a particular outcome for \(X\) given the total sum is 2. The process involves calculating the joint probability for each \(k\). In this case:

• Calculate \(P(X=k)\) and \(P(Y=2-k)\) using their respective Poisson distributions.
• Find the joint probabilities \(P(X=k, Y=2-k)\).
• Divide each joint probability by \(P(X+Y=2)\) to find the conditional probabilities.

This method provides insights into how likely each \(X=k\) is when you know the sum. Understanding these probabilities helps in various real-world scenarios where outcomes depend on certain conditions.